Question

The angular momentum of an electron in a Bohr orbit of $H$ atom is $4.4\times {10}^{-34}kg{m}^2{s}^{-1}$ . The wavelength of the spectral line emitted when the electron falls from this level to the next lower level is $x\times {10}^{-4}c{m}^{-1}$. The value of $x$ is
(Given: $R_H=1.1\times {10}^{5}c{m}^{-1}$)
(take: $\pi =3,\text{and Plank's constant}$ = $6.6\times {10}^{-34}$)

Answer 1

Given angular momentum in a orbit as:
$mvr=\frac{nh}{2\pi }$

$\therefore \frac{nh}{2\pi }=4.4\times {10}^{-34}kg{m}^2{s}^{-1}$

or $n=\frac{4.4\times {10}^{-34}\times 2\times 3}{6.6\times {10}^{-34}}$
$n=4$

using Rydberg's formula:
$\frac{1}{\lambda }=R_H[\frac{1}{n_1^2}-\frac{1}{n_1^2}]$
putting $n_1=3$ and $n_2=4$ we get,
$\frac{1}{\lambda }=1.1\times {10}^5[\frac{1}{3^2}-\frac{1}{4^2}]$
gives, $\lambda =1.87\times {10}^{-4}cm$